What is the extreme f(x,y)=xy-x^2-y^2-2x-2y+4 ?

The extreme f(x,y)=xy-x^2-y^2-2x-2y+4 is Maximum(-2,-2)

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extreme f(x,y)=xy-x^2-y^2-2x-2y+4

extreme points

f (x, y) = x y - x^{2} - y^{2} - 2 x - 2 y + 4

Maximum (- 2, - 2)

Solution steps

Find the critical points:

(- 2, - 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 2

D (x, y) = 3

Check the sign of at each critical point

Check critical point

(- 2, - 2) :

Maximum

Maximum (- 2, - 2)

f (x) = \frac{x ^{2} - 2 x + 1}{x + 1}

f (x) = x \cdot ln (x)

f (x) = (cos (x) - 0.5)^{2}

f (x) = - x^{3} + 6 x^{2} - 10 x + 4

f (x) = 4 x^{3} - 3 x^{2} - 6 x + 3

What is the extreme f(x,y)=xy-x^2-y^2-2x-2y+4 ?
The extreme f(x,y)=xy-x^2-y^2-2x-2y+4 is Maximum(-2,-2)